Chapter 5 Present Worth Analysis EGN 3615 ENGINEERING ECONOMICS WITH SOCIAL AND GLOBAL IMPLICATIONS.

Chapter 5Present Worth Analysis

EGN 3615 ENGINEERING ECONOMICS WITH SOCIAL AND GLOBAL

IMPLICATIONS

Three Economic Analysis MethodsThere are three major economic analysis

techniques:

Present Worth AnalysisAnnual Cash Flow AnalysisRate of Return Analysis

This chapter discusses the first techniques

2

Chapter ContentsEconomic CriteriaConsidering Project LifeNet Present WorthApplying Present Worth Techniques

Useful Lives Equal the Analysis PeriodUseful Lives Different from the Analysis

PeriodInfinite Analysis Period: Capitalized CostMultiple Alternatives

Spreadsheet Solution3

Economic CriteriaDepending on situation, the economic

criterion should be chosen from one of the following 3:

Situation CriterionNeither input nor output fixed

Maximize (Output – Input)

Fixed input Maximize outputFixed output Minimize input

4Engineering Economics

Analysis PeriodSpecific time period, same for each

alternative, called the analysis period, planning horizon, or project life

Three different analysis-period situations may be considered:1. All alternatives have the same useful life: Set

it as the analysis period.2. Alternatives have different useful lives: Let

the analysis period equal the least common multiple, or some realistic time (based on needs).

3. Infinite analysis period, n=∞ 5Engineering Economics

Net Present Worth (NPW or PW)Here is the basic NPW formula:

PW = PW of benefits – PW of cost

Costs

of

rth Present wo

-

Benefits

of

rth Present wo

PWor NPW

WORTH

PRESENT NET


Present Worth TechniquesMutually exclusive alternatives: Resolve their consequences to the present

time. Situation Criterion

Neither input nor output fixed

Maximize net present worth

Amount of money or other input resources are fixed

Maximize present worth of benefits or other outputs

Fixed task, benefit, or other outputs

Minimize present worth of costs or other inputs


Present Worth—Equal Useful LivesExample: Consider two mechanical devices to

install to reduce cost. Expected costs and benefits of machines are shown in the following table for each device. If interest rate is 6%, which device should be purchased?DEVICE COS

TCOST SAVING USEFUL

LIFEDEVICE A

$1000

$300 Annually5 year

DEVICE B

$1350

$300 The first year and increase $50 annually

5 year8Engineering Economics

Example Continues

263.6

(4.212) 300 1000

(P/A,6%,5) 3001000PW A

0

1

2 3

4 5

A=$300

P= $1000

i=6%


Example Continues

310.3

(7.934) 50 (4.212) 300 1350

(P/G,6%,5) 50 (P/A,6%,5) 300 1350 PW B

0

1

2 3

4

$300

P= $1350

i=6%

$350

$400

$450

$500

5


Example Continues

310.3 PW

263.6 PW

B

A

0

1

2 3

4 5

A=$300

P= $1000

i=6%

DEVICE B has the larger present worth & is the preferred alternative

0

1

2 3

4

$300

P= $1350

i=6%

$350

$400

$450

$500

5


Work 5-4

Example: Consider two investments with expected costs and benefits shown below for each investment. If investments have lives equal to the 5-year analysis period, which one should be selected at 10% interest rate?

Investment

Cost Benefit Useful

Life

Salvage Value (End of Useful

Life) Investment 1

$2000

$450 Annually

5 year $100

Investment 2

$3000

$600 Annually

5 year $700

Present Worth—Equal Useful Lives


Example Continues

96.231

] (0.6209) 100200[(3.791) 450

10%,5)] (P/F, 100200[5) 10%, (P/A, 450

Costs ofPW Benefits ofPW

:1 Investment


Example Continues

290.77

] (0.6209) 7003000[(3.791) 600

] 10%,5) (P/F, 7003000[5) 10%, 600(P/A,

Costs ofPW Benefits ofPW

:2 Investment


Example Continues

290.77

(0.6209)] 7003000[(3.791) 600

10%,5)] (P/F, 7003000[5) 10%, 600(P/A,Costs ofPW - Benefits ofPW

:2 Investment

96.231

] (0.6209) 100200[(3.791) 450

] 10%,5) (P/F, 100200[5) 10%, (P/A, 450Costs ofPW - Benefits ofPW

:1 Investment

Salvage value is considered as another positive cash flow. Since criterion is to maximize PW (= present worth of benefits – present worth of costs), the preferred alterative is INVESTMENT1


EQUIPMENT

COSTSALVAGE

VALUEUSEFUL

LIFEEQUIPMENT

A$1500 $200 5 year

EQUIPMENT B

$1600 $350 10 year

Example: Consider two new equipments to perform desired level of (fixed) output. expected costs and benefits of machines are shown in the below table for each equipment. If interest rate is 6%, which equipment should be purchased?

Alternatives with different Useful Lives


Example ContinuesOne method to select an analysis period is

the least common multiple of useful lives.

0

1

2 3

4 5

$200

$1500

6 7 8 9 10

$1500

$200Replacement

Equipment A Investment

Original Equipment A Investment

EQUIPMENT A

81. 359,2$

) 5584.0( 200 ) 7473.0( 1300 1500

10) 6%, (P/F, 2005) 6%, (P/F, 200) (1500 1500

cost ofPW


Question Continues$350

0

1

2 3

4 5

$1600

6 7 8 9 10

Original Equipment B Investment

EQUIPMENT B

1,404.56$

(0.5584) 350 1600

10) 6%, (P/F, 350 1600

cost ofPW


Question Continues

$1,404.56

(0.5584) 325 1600

10) 6%, (P/F, 325 1600cost ofPW

81. 359,2$

) 5584.0( 200 ) 7473.0( 1300 1500

10) 6%, (P/F, 200

5) 6%, (P/F, 200)- (1500 1500cost ofPW

For fixed output of 10 years of service of equipments, Equipment B is preferred because it has a smaller cost.

EQUIPMENT A

EQUIPMENT B


Example: Consider two alternative production machines with expected initial costs & salvage values shown below. If interest rate is 10%, compare these alternatives over a (suitable) 10-year analysis period (by using the present worth method)?

MACHINEINITIAL

COST

Salvage Value at the End

Of Useful Life

Terminal Value at the end of 10-

year analysis period

USEFUL LIFE

MACHINE A

$40,000

$8,000 $15,000 7 year

MACHINE B

$65,000

$10,000 $10,000 13 year

Present Worth-Useful Lives are Different from the Analysis Period


Example Continues

$40,000

0

1

2 3

4 5

$8,000

6 7 8 9 10

$15,000

7-year life

MACHINE A

11 12 13 14

$40,0007-year life

50,639.90$

) 0.3855( 15,000) 5132.0( 000,3240,000

10) 10%, (P/F, 15,000

7) 10%, F,8,000)(P/ (40,00040,000

cost ofPW


Example Continues

0

1

2 3

4 5 6 7 8 9 10

$10,000

$65,000

MACHINE B

11 12 13 14

13-year life

61,145.0$

0.3855) ( 10,000 000,56

10) 10%, (P/F, 10,000 000,56

cost ofPW


Example Continues

$61,145.00

0.3855) ( 10,000 000,56

10) 10%, (P/F, 10,000 000,56cost ofPW

$50,639.90

) 0.3855( 15,000 ) 5132.0( 000,32 40,000

10) 10%, (P/F, 15,000

7) 10%, (P/F, 8,000)- (40,000 40,000cost ofPW

For fixed output of 10 years of service of equipments, Machine A is preferred because it has a smaller cost.

MACHINE B

MACHINE A


Infinite Analysis Period (Capitalized Cost)

Capitalized cost is the present sum of money that is set aside now at a given interest rate to yield the funds (future interest earned) required to provide the service indefinitely.

(5-2)

i

A PCost dCapitalize


Infinite Analysis Period (Capitalized Cost)

Example: How much should one set aside to pay $1000 per year for maintenance on an equipment if interest rate is 2.5% per year and the equipment is kept in service indefinitely (perpetual maintenance)?

000,40$0.025

1000P

(i) rateInterest

(A)nt disburseme AnnualP Cost, dCapitalize


Multiple (3+) AlternativesQuestion: Cash flows (costs and incomes) for

three pieces of construction equipments are shown below. For 10% interest rate, which alternative should be selected?Year Equipment

1Equipment

2Equipment

3

0 -$2000 -$1500 -$3000

1 +1000 +700 +500

2 +850 +300 +500

3 +700 +300 +550

4 +550 +300 +600

5 +400 +300 +650

6 +400 +400 +700

7 +400 +500 +500

8 +400 +600 +50026Engineering Economics

Question Continues

$2000

0

1

2 3

4 5

$400

6 7 8

$400

EQUIPMENT 1

$1000

$850

$700

$550

$400

$400

17.379,1$200017.3379rthPresent WoNet

2000 Cost ofPW

3379.17150(4.378)

)170.3(600)335.5(400

4) 10%, 150(P/G,

4) 10%, 600(P/A,8) 10%, 400(P/A,Benefits ofPW


Question Continues

0

1

2 3

4 5

$300

$1500

6 7 8

$600

EQUIPMENT 2

$700

$300

$300

$300

$400

$500

15.763$150015.2263rthPresent WoNet

1500 Cost ofPW

15.2263(0.6830) 100(4.378)

)9091.0(400)335.5(300

4) 10%, 4)(P/F, 10%, 100(P/G,

1) 10%, 300)(P/F,-007(

8) 10%, 300(P/A,Benefits ofPW


Question Continues

0

1

2 3

4 5

$650

$3000

6 7 8

$500

EQUIPMENT 3

$500

$500

$550

$600

$700

$500

64.20$300036.2979rthPresent WoNet

3000 Cost ofPW

36.2979(0.9091)100(6.862)

)335.5(500

1) 10%, 5)(P/F, 10%, 50(P/G,

8) 10%, 500(P/A,Benefits ofPW

To maximize NPW, choose EQUIPMENT 1 29Engineering Economics

Question Continues (MS EXCEL)


Use function: npv(rate, value range) - Return the net present value of a series of future cash flows “value range” at interest “rate”/period.

rate = interest rate per period

value range = the cash flow values

Question Continues (MS EXCEL)Year Equipment 1 Equipment2 Equipment3

0 ($2,000) ($1,500) ($3,000)

1 1000 700 500

2 850 300 500

3 700 300 550

4 550 300 600

5 400 300 650

6 400 400 700

7 400 500 500

8 400 600 500

Interest For Equip 1: NPW=NPV(A12,B3:B10)+B2

10% $1,379.17 $763.15 ($20.64)


Problem 5-15


Solution i = 12%P = $980,000 purchase costF = $20,000 salvage value after

13 yearsA = $200,000 annual benefit for 13

years

PW = –P + A(P/A, 0.12, 13) + F(P/F, 0.12, 13) = –980000 + 200000(6.424) + 20000(0.2292) = $309,384As PW > 0, purchase the machine.

Or using MS EXCELPW = -P + pv(0.12, 13, -200000, -20000) = $309,293.17Terms A(P/A, 0.12, 13) and F(P/F, 0.12, 13) are combined!

Problem 5-23


Solution i = 18%/12 = 1.5% per monthA = $500 payment/monthn = 36 paymentsP = ? price of a car she can

afford

P = A(P/A, 0.015, 36) = 500(27.661) = $13,831

What is P, if r = 6%? i = 6%/12 = 0.5%P = pv(0.005, 36, -500) = $16,435.51

Do Problems 5-24, 5-25, 5-26!

Problem 5-41


Outputs: 2000 lines for years 1~104000 lines for years 21~30

i = 10% per year, cables last for at least 30 yrsOption 1: 1 cable with capacity of 4000 lines

Cost: $200k with $15k annual maintenance costOption 2: 1 cable with capacity of 2000 lines now

1 cable with capacity of 2000 lines in 10 years

Cost: $150k with $10k maintenance cost/year/cable

(a) Which option to choose?(b) Will answer to (a) change if 2000 additional lines are

needed in 5 years, instead of 10 years?

Problem 5-41


Solution(a)Present worth of cost for option 1PW 0f cost = $200k + $15k(P/A, 10%, 30)

= $341,400

Present worth of cost for option 2:PW of cost = $150k + $10k(P/A, 10%, 30)

+ $150k(P/F, 0.1, 10) + $10k(P/A, 0.1, 20)(P/F, 0.1, 10)

= $334,900

Select option 2, as it has a smaller PW of cost.

Problem 5-41


Solution(b)Cost for option 1 will not change.PW 0f cost = $341,400

Present worth of cost for option 2:PW of cost = $150k + $10k(P/A, 10%, 30)

+ $150k(P/F, 0.1, 5) + $10k(P/A, 0.1, 25)(P/F, 0.1, 5)

= $394,300

Therefore, the answer will change to option 1.

End of Chapter 5


Chapter 5 Present Worth Analysis EGN 3615 ENGINEERING ECONOMICS WITH SOCIAL AND GLOBAL IMPLICATIONS.

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Transcript of Chapter 5 Present Worth Analysis EGN 3615 ENGINEERING ECONOMICS WITH SOCIAL AND GLOBAL IMPLICATIONS.